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Question #44714

Difficulty: 8/10 | 🟢 Free

The Question

A saturated aqueous solution of cerium(III) sulfate, \ce{Ce2(SO4)3}, is prepared at 320 K320 \text{ K} in 1 kg1 \text{ kg} of water. The excess un-dissolved solid is rapidly filtered off at 320 K320 \text{ K}. The filtrate is then cooled to determine its freezing point. Assuming ideal behavior and complete dissociation of the salt, what is the freezing point depression (ΔTf\Delta T_f) of this filtrate?

Given: Solubility of \ce{Ce2(SO4)3} at 300 K=2.0×10−2 mol kg−1300 \text{ K} = 2.0 \times 10^{-2} \text{ mol kg}^{-1} ΔHsol∘=−138.28 kJ mol−1\Delta H^\circ_{\text{sol}} = -138.28 \text{ kJ mol}^{-1} (assume independent of temperature) Kf for water=1.86 K kg mol−1K_f \text{ for water} = 1.86 \text{ K kg mol}^{-1} R=8.314 J K−1mol−1R = 8.314 \text{ J K}^{-1} \text{mol}^{-1} ln⁡2=0.693\ln 2 = 0.693

Options

  • A.

    0.093 K

  • B.

    0.372 K

  • C.

    0.186 K

  • D.

    0.0186 K

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Solubility of Solids in Liquids|Gibbs Energy and Spontaneity|Freezing Point Depression (Pure Solvent)

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Le Chatelier's Principle, the Van't Hoff equation, and colligative properties.

The 'Chemical Exception' is retrograde solubility: because the dissolution of \ce{Ce2(SO4)3} is exothermic, its solubility decreases at higher temperatures.

Students commonly assume solubility always increases with temperature, leading them to incorrectly double the solubility instead of halving it (Trap B).

Furthermore, the 'Hidden Variable' is the phase behavior during cooling: if the dissolution were endothermic, cooling the solution would cause precipitation, altering the molality before freezing.

Because it is exothermic, cooling increases the solubility limit, ensuring the solution becomes unsaturated and the molality remains perfectly constant at the value established at 320 K.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterSolutions - Class 12
ConceptSolubility of Solids in Liquids
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree